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How to find term of AP when other two terms of AP are given?

CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Question 8



8.   An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

 

 

Solution:

 

An AP consists of 50 terms and the 50th term is equal to 106.

 

It is also given that a_{3}=12.           {a_3 is the 3rd term of AP}

 

 

Using formula {{a}_{n}}=a+(n-1)d,   to find nth term of arithmetic progression, we get

 

a_{50} = a + (50-1)d      And   a_3 =a + (3-1)d

 

\Rightarrow 106 = a + (50-1)d     And   12 = a + (3-1)d

 

\Rightarrow 106 = a + 49d    And  12 = a + 2d

 

These are equations consisting of two variables. Lets solve them using substitution method.

Using equation 106 = a + 49d, we get   a = 106-49d

 

 

Putting value of a in the equation 12 = a + 2d, we get

12 = 106-49d+2d

 

\Rightarrow 47d = 94

 

\Rightarrow d = \frac{94}{47} =2

 

Putting value of d in the equation, a = 106-49d, we get

a = 106 -49 (2) = 106 - 98 = 8

 

Therefore, First term = a = 8 and Common difference = d = 2

 

To find 29th term, we use formula {{a}_{n}}=a+(n-1)d which is used to find nth term of arithmetic progression, we get

{{a}_{29}}=8+(29-1)2 = 8 + 56 = 64

 

Therefore, 29th term of AP is equal to 64.

 



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